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Number of integral values

  1. Nov 21, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Number of integral values of m for the equation [itex]\sin x - \sqrt{3} \cos x = \dfrac{4m-6}{4-m}[/itex] to have a solution is

    3. The attempt at a solution

    I started by plotting the graph of sinx and √3 cosx. Now if I shift the graph of √3 cosx to √3 units above or below its current position I can still have a solution. In other way 4m-6/4-m can be seen as a numerical constant that determines the shift in the graph of √3 cosx. Imposing the inequality

    [itex]- \sqrt{3} \leq \dfrac{4m-6}{4-m} \leq \sqrt{3} [/itex]

    and solving, I get a different answer.
     
  2. jcsd
  3. Nov 21, 2013 #2

    HallsofIvy

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    You might find it easier to use a trig identity to reduce to a single sine.

    sin(a+ b)= cos(a)sin(b)+ sin(a)cos(b). Taking b= x, we would like to have cos(a)= 1, [itex]sin(a)= \sqrt{3}[/itex]. Of course, that is impossible because [itex]\sqrt{3}[/itex] is larger than 1. But we can "adjust" that. [itex]\sqrt{1^2+ 3^2}= \sqrt{10}[/itex] so [itex]sin(x)- \sqrt{3}cos(x)= \sqrt{10}\left((1/\sqrt{10} sin(x)+ (-\sqrt{3}/\sqrt{10}) cos(x)\right)[/itex]. Now we look for a such that [itex]cos(a)= 1/\sqrt{10}[/itex] and [itex]sin(a)= -\sqrt{3}/\sqrt{10}[/itex]. That is a= -1.25 radians.

    That is, [tex]sin(3)- \sqrt{3}cos(x)= \sqrt{10}sin(x- 1.25)= \dfrac{4m- 6}{4- m}[/tex].

    Since sine is always between -1 and 1, we must have [tex]-\sqrt{10}\le\dfrac{4m-6}{4-m}\le \sqrt{10}[/tex].
     
  4. Nov 21, 2013 #3

    Ray Vickson

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    I get [tex]f(x) = \frac{1}{2} \sin(x - \pi/3).[/tex] Of course ## 1^2 + \sqrt{3}^2 = 4##, not 10.

    Anyway, could we have not left some of the work for the OP to do?
     
  5. Nov 21, 2013 #4

    utkarshakash

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    I think f(x)=2sin(x-∏/3). Doing that way I get the correct answer. But can you please tell what is wrong in my original solution?
     
  6. Nov 21, 2013 #5

    Ray Vickson

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    I could not figure out what you were trying to do.
     
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